In a refrigerator, the low temperature coil of evaporator is at `-23^@C` and the compressed gas in the condenser has a temperature of `77^@C`. The amount of electrical energy spent in freezing 1 kg of water at `0^@C` is

Options

A. `134400 J`
B. `1344 J`
C. `80000 J`
D. `3200 J`

Answer

To solve the problem of calculating the amount of electrical energy spent in freezing 1 kg of water at 0°C in a refrigerator with the given parameters, we can follow these steps:

### Step 1: Calculate the heat required to freeze 1 kg of water
The heat required to freeze water can be calculated using the formula:
\[ Q_L = m \cdot L_f \]
where:
- \( Q_L \) = heat required to freeze the water
- \( m \) = mass of the water = 1 kg = 1000 grams
- \( L_f \) = latent heat of fusion of water = 80 cal/g

Substituting the values:
\[ Q_L = 1000 \, \text{g} \cdot 80 \, \text{cal/g} = 80000 \, \text{cal} \]

### Step 2: Convert calories to joules
To convert calories to joules, we use the conversion factor \( 1 \, \text{cal} = 4.2 \, \text{J} \):
\[ Q_L = 80000 \, \text{cal} \cdot 4.2 \, \text{J/cal} = 336000 \, \text{J} \]

### Step 3: Determine the temperatures in Kelvin
We need to convert the temperatures from Celsius to Kelvin:
- Lower temperature (evaporator) = \(-23°C + 273 = 250 \, K\)
- Higher temperature (condenser) = \(77°C + 273 = 350 \, K\)

### Step 4: Calculate the efficiency of the refrigerator
The efficiency (COP) of the refrigerator can be calculated using the formula:
\[ \text{COP} = \frac{T_L}{T_H - T_L} \]
Substituting the values:
\[ \text{COP} = \frac{250}{350 - 250} = \frac{250}{100} = 2.5 \]

### Step 5: Relate heat absorbed and work done
Using the relationship:
\[ Q_H = Q_L + W \]
where \( W \) is the work done. We can express \( W \) in terms of \( Q_L \):
\[ W = Q_H - Q_L \]

### Step 6: Calculate the heat rejected to the higher temperature body
Using the efficiency:
\[ W = \frac{Q_L}{\text{COP}} = \frac{336000 \, \text{J}}{2.5} = 134400 \, \text{J} \]

### Final Answer
Thus, the amount of electrical energy spent in freezing 1 kg of water at 0°C is:
\[ \boxed{134400 \, \text{J}} \]